1952 AHSME Problems/Problem 22

Problem

On hypotenuse $AB$ of a right triangle $ABC$ a second right triangle $ABD$ is constructed with hypotenuse $AB$ . If $\overline{BC}=1$ , $\overline{AC}=b$ , and $\overline{AD}=2$ , then $\overline{BD}$ equals:

$\textbf{(A) \ }\sqrt{b^2+1} \qquad \textbf{(B) \ }\sqrt{b^2-3} \qquad \textbf{(C) \ }\sqrt{b^2+1}+2 \qquad$

$\textbf{(D) \ }b^2+5 \qquad \textbf{(E) \ }\sqrt{b^2+3}$

Solution

$[asy] size(150); import olympiad; pair A,B,C,D; A=(1,3); B=origin; C=(1,0); D=sqrt(6)*dir(aTan(2/sqrt(6))+aTan(9)); draw(A--C--B--A--D--B); markscalefactor=0.02; draw(rightanglemark(A,C,B)); draw(rightanglemark(A,D,B)); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$1$",B--C,S); label("$x$",D--B,W); label("$2$",D--A,N); label("$b$",A--C,E); [/asy]$ Let $\overline{BD}=x$ . According to the Pythagorean Theorem, $x^2+4=b^2+1$ . Hence, $x=\boxed{\textbf{(B)}\ \sqrt{b^2-3}}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1952 AHSME Problems/Problem 22

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